Cast variable to integer or null if not able to

Question

In PHP I have 3 variables set like this...

$myVariable1  = '2345';
$myVariable2  = 4433;
$myVariable3  = 'test';

I am trying to ensure that these are all integers and if not then I would like to set them as blank. I have read up on is_numeric but am not sure if this is the correct function to use.

if (!is_numeric($myVariable1)) {
    $myVariable1 = 2345;
}

if (is_numeric($myVariable2)) {
    $myVariable2 = 4433;
}

if (!is_numeric($myVariable3)) {
    $myVariable3 = NULL;
}

Is there a way to do this automatically so it will attempt to cast to an integer or set NULL if not able to?

If you're looking for integers specifically then use [is_int](https://www.php.net/manual/en/function.is-int.php). is_numeric will allow a much wider variety of numbers (see the [docs](https://www.php.net/manual/en/function.is-numeric.php)). — ADyson, Jan 31 '20 at 09:23
is_int looks good, but is there a way to attempt to cast it as an integer as well? — fightstarr20, Jan 31 '20 at 09:25
Why would you want to cast to integer as blank? You shouldn't blacklist values, as this renders your code outdated very quickly. Best to check if values ARE ints, and then run your code on that int if they are. Otherwise, don't run that code. — cmprogram, Jan 31 '20 at 09:27
Did you google, for example "php cast to int"?? You'll get plenty of suggestions. https://stackoverflow.com/a/8529687/5947043 is useful info. You could read https://www.php.net/manual/en/language.types.type-juggling.php and https://www.php.net/manual/en/language.types.integer.php#language.types.integer.casting as well. PHP is quite flexible abouts types - as long as it passes the is_int test you probably won't need to explicitly cast it most of the time. — ADyson, Jan 31 '20 at 09:27
Casting `'test'` into an int would only get you `0`, not `NULL`. — 04FS, Jan 31 '20 at 09:28

Aksen P · Answer 1 · 2020-01-31T09:33:27.200

1

You can do something like this:

$myVariable1  = '2345';
$myVariable2  = 4433;
$myVariable3  = 'test';
$myVariable4  = '1345.33';

function makeInteger($val){
    return is_numeric($val) ? (int)$val : null;
}

for($i=1;$i<=4;$i++){ 
    echo makeInteger(${"myVariable".$i}).PHP_EOL; 
}

Output:

Demo

Note: casting (int) will remove float part of the variable.

edited Jan 31 '20 at 09:33

answered Jan 31 '20 at 09:30

Aksen P

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score 1 · Accepted Answer · answered Jan 31 '20 at 09:32

Try this

$a = '1234';
$b = '3333';
$c = 'test';

list($a1, $b1, $c1) = array_map(function($elem) { return is_numeric($elem) ? intval($elem) : null; }, [$a, $b, $c]);

Result

print_r([$a1, $b1, $c1]);
Array
(
    [0] => 1234
    [1] => 3333
    [2] =>
)

Cast variable to integer or null if not able to

2 Answers2